HDU 1402 A * B Problem Plus（FFT）

Problem Description

Calculate A * B.

 

Input

Each line will contain two integers A and B. Process to end of file.

Note: the length of each integer will not exceed 50000.

 

Output

For each case, output A * B in one line.

 

题目大意：求A * B。

思路：快速傅里叶变换的模板题，偷的模板……也不知道使用姿势对不对QAQ。

 

代码（250MS）：（Update：2014年11月16日）

 1 #include <cmath>

 2 #include <algorithm>

 3 #include <cstdio>

 4 #include <iostream>

 5 #include <cstring>

 6 #include <complex>

 7 using namespace std;

 8 typedef complex<double> Complex;

 9 const double PI = acos(-1);

10 

11 void fft_prepare(int maxn, Complex *&e) {

12     e = new Complex[2 * maxn - 1];

13     e += maxn - 1;

14     e[0] = 1;

15     for (int i = 1; i < maxn; i <<= 1)

16         e[i] = Complex(cos(2 * PI * i / maxn), sin(2 * PI * i / maxn));

17     for (int i = 3; i < maxn; i++)

18         if ((i & -i) != i) e[i] = e[i - (i & -i)] * e[i & -i];

19     for (int i = 1; i < maxn; i++) e[-i] = e[maxn - i];

20 }

21 /* f = 1: dft; f = -1: idft */

22 void dft(Complex *a, int N, int f, Complex *e, int maxn) {

23     int d = maxn / N * f;

24     Complex x;

25     for (int n = N, m; m = n / 2, m >= 1; n = m, d *= 2)

26         for (int i = 0; i < m; i++)

27             for (int j = i; j < N; j += n)

28                 x = a[j] - a[j + m], a[j] += a[j + m], a[j + m] = x * e[d * i];

29     for (int i = 0, j = 1; j < N - 1; j++) {

30         for (int k = N / 2; k > (i ^= k); k /= 2);

31         if (j < i) swap(a[i], a[j]);

32     }

33 }

34 

35 const int MAXN = 131072;

36 Complex x1[MAXN], x2[MAXN];

37 char s1[MAXN / 2], s2[MAXN / 2];

38 int sum[MAXN];

39 

40 int main() {

41     Complex* e = 0;

42     fft_prepare(MAXN, e);

43     while(scanf("%s%s",s1,s2) != EOF) {

44         int n1 = strlen(s1);

45         int n2 = strlen(s2);

46         int n = 1;

47         while(n < n1 * 2 || n < n2 * 2) n <<= 1;

48         for(int i = 0; i < n; ++i) {

49             x1[i] = i < n1 ? s1[n1 - 1 - i] - '0' : 0;

50             x2[i] = i < n2 ? s2[n2 - 1 - i] - '0' : 0;

51         }

52 

53         dft(x1, n, 1, e, MAXN);

54         dft(x2, n, 1, e, MAXN);

55         for(int i = 0; i < n; ++i) x1[i] = x1[i] * x2[i];

56         dft(x1, n, -1, e, MAXN);

57         for(int i = 0; i < n; ++i) x1[i] /= n;

58 

59         for(int i = 0; i < n; ++i) sum[i] = round(x1[i].real());

60         for(int i = 0; i < n; ++i) {

61             sum[i + 1] += sum[i] / 10;

62             sum[i] %= 10;

63         }

64 

65         n = n1 + n2 - 1;

66         while(sum[n] <= 0 && n > 0) --n;

67         for(int i = n; i >= 0;i--) printf("%d", sum[i]);

68         puts("");

69     }

70 }

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